Question: Let $g(x)=\sqrt[4]{x^3}$. $g'(1)=$
Explanation: The strategy We can first rewrite $g(x)$ as a rational power of $x$. Then, the derivative of $g$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have $g'(x)$, we can plug $x=1$ into it to find $g'(1)$. Rewriting the radical as a rational power $g(x)=\sqrt[4]{x^3}=x^{^{\frac{3}{4}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{3}{4}}}\right) \\\\ &=\dfrac{3}{4}x^{^{\frac{3}{4}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac{3}{4}x^{^{-\frac{1}{4}}} \end{aligned}$ Evaluating $g'(x)$ So we found that $g'(x)=\dfrac{3}{4}x^{^{-\frac{1}{4}}}$, which can also be written as $\dfrac{3}{4\sqrt[4]{x}}$. Now let's plug ${x=1}$ : $\begin{aligned} \dfrac{3}{4\sqrt[4]{{1}}}&=\dfrac{3}{4\cdot 1} \\\\ &=\dfrac34 \end{aligned}$ In conclusion, $g'(1)=\dfrac{3}{4}$.